STRUCTURE OF Zr-90, Zr-91, Zr-92 AND Zr-94
By Prof. Lefteris Kaliambos (Natural Philosopher in mew Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of zirconium ( Zr) Using the diagram of Zr with 40 deuterons you see that the 36 deuterons from p1n1 to p36n36 have the structure of a core representing a complete shape of six horizontal planes . Loking on the top view of the first plane one concludes that each plane has 6 deuterons and 2 extra neutrons (n) of weak bonds. That is, the core of 36 deuterons has 12 extra neutrons (n) of weak bonds. Note that all nucleons of the 6 planes have spin S= 0 because the first plane is characterized by positive spins, while the sixth plane has nucleons of negative spin. In general to reveal the structure of zirconium with 40 deuterons ( even number of high symmetry) we add the square of p37n37 and n39p39 with negative spins under the n2p2 and p1n1. Thus the symmetrical square of p38n38 and n40p40 with positive spins is over the n11p11 and p12n12. Here the p38 and p40 can receive 4 extra neutrons n of positive spins giving strong bonds, while the p37 and p39 can receive 4 extra neutrons n of negative spins giving strong bonds. For example the p37 with p34 makes a blank position for receiving an extra neutron n which makes the n-p34 strong bond and the n-p37 weak bond. .Naturally occurring zirconium (Zr) is composed of stable isotopes (of which one may in the future be found radioactive), and one very long-lived radioisotope (96 Zr). STRUCTURE OF Zr-90, Zr-92 AND Zr-94 WITH S = 0 The Zr-90 is based on Zr-80 with S =0 which has 40 deuterons with S=0. In this structure one adds 10 extra neutrons . Since the structure of Zr has 8n one concludes that 10 = 8n +2(n) with opposite spins. In the same way the Zr-92 has 8n + 4(n) = 12 which is the number of extra neutrons. Also the Zr-94 with S =0 has 40 deuterons of opposite spin and 14 extra neutrons of opposite spin That is 14 = 8n +6(n) In other words all the above nuclides have a stable structure because the number of the extra neutrons n of strong bonds is greater than the number of extra neutrons (n) with weak bonds. STRUCTURE OF Zr-91 WITH S = +5/2 ' Here the new symmetrical arrangements lead to the stable structure when the deuteron p37n37 changes the spin from S=-1 to S =+1 in order to fill the blank position in front of p38n38. Thus the change of spin gives S =+2 Under this condition one concludes that Zr-91 will receive 11 extra neutron from which 6 extra neutrons will have positive spin and five neutrons will have negative spin. That is S = +2 + 6(+1/2) + 5(-1/2) = + 5/2 ' ''' '''DIAGRAM OF Zr 80 WITH 40 DEUTERONS , HAVING ALSO 8 EXTRA n AND 12 EXTRA (n) Here the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Also the 12 extra neutrons (n) with weak bonds of six planes and the 4 extra neutrons n of strong bonds of n near the p37, p38, p39 and p40 are not shown. You can see only 4 extra neutrons n of strong bonds existing under the p21 and p22 and over the p31 and p32. ' ' n40.........p40.......n ' n........p38..........n38 Horizontal squarewith n ' ' n31………p12........n12........p32' ' p31....... n11.........p11…… n32 Sixth horizontal plane' ' p29....... n10.........p10…….n30' ' n29………..p9..........n9 …….p30 Fifth horizontal plane' ' n27.........p8..........n8...........p28' ' p27.........n7..........p7........n28 Fourth horizontal plane' ' p25.........n6.........p6.............n26' ' n25……….p5........n5……p26 Third horizontal plane' ' n23………p4........n4………p24' ' p23……..n3……p3………n24 Second horizontal plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First horizontal plane' ' n'.........p37 ......n37 ' ' n39......p39.........n Horizontal square with n ' ' ' '''TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ''' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' Category:Fundamental physics concepts